Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $t = \dfrac{k^2 + 6k - 16}{-9k - 45} \times \dfrac{k + 5}{-k + 2} $
Explanation: First factor the quadratic. $t = \dfrac{(k - 2)(k + 8)}{-9k - 45} \times \dfrac{k + 5}{-k + 2} $ Then factor out any other terms. $t = \dfrac{(k - 2)(k + 8)}{-9(k + 5)} \times \dfrac{k + 5}{-(k - 2)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (k - 2)(k + 8) \times (k + 5) } { -9(k + 5) \times -(k - 2) } $ $t = \dfrac{ (k - 2)(k + 8)(k + 5)}{ 9(k + 5)(k - 2)} $ Notice that $(k + 5)$ and $(k - 2)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ \cancel{(k - 2)}(k + 8)(k + 5)}{ 9(k + 5)\cancel{(k - 2)}} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $t = \dfrac{ \cancel{(k - 2)}(k + 8)\cancel{(k + 5)}}{ 9\cancel{(k + 5)}\cancel{(k - 2)}} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $t = \dfrac{k + 8}{9} ; \space k \neq 2 ; \space k \neq -5 $